Gravitation
Gravitation for JEE Main & Advanced
Newton's Law of Gravitation and Gravitational Field
Newton's Law, Gravitational Force and Acceleration due to GravityTopic 1
Newton's Law of Gravitation: Every particle attracts every other particle with a force: $$F = \frac{Gm_1m_2}{r^2}$$ where $G = 6.674 \times 10^{-11}$ N·m²/kg² is the universal gravitational constant. The force is always attractive, acts along the line joining the centres, and obeys the inverse-square law.
Acceleration due to gravity at Earth's surface: $g = \dfrac{GM_e}{R_e^2} \approx 9.8$ m/s². Mass of Earth: $M_e = gR_e^2/G$.
Variation of $g$:
| Location | Formula | Effect |
|---|---|---|
| Height $h$ above surface | $g_h = g\left(\dfrac{R}{R+h}\right)^2 \approx g\left(1 - \dfrac{2h}{R}\right)$ for $h \ll R$ | Decreases |
| Depth $d$ below surface | $g_d = g\left(1 - \dfrac{d}{R}\right)$ | Decreases (zero at centre) |
| Latitude $\lambda$ | $g_\lambda = g - R\omega^2\cos^2\lambda$ | Max at poles, min at equator |
| Rotation effect | $\Delta g = R\omega^2\cos^2\lambda$ | Reduces $g$ |
Key Numbers: $R_e \approx 6400$ km; $M_e \approx 6 \times 10^{24}$ kg; $g_{\text{pole}} - g_{\text{equator}} \approx 0.034$ m/s².
Find the height above Earth's surface where $g$ becomes $1/4$ of its surface value. ($R = 6400$ km)
Show solution
$$\frac{g_h}{g} = \left(\frac{R}{R+h}\right)^2 = \frac{1}{4}$$ $$\frac{R}{R+h} = \frac{1}{2} \implies R+h = 2R \implies h = R = 6400 \text{ km}$$
Final Answer: $h = 6400$ km.
At what depth below Earth's surface does the value of $g$ become $50\%$ of its surface value?
Show solution
$$g_d = g\left(1 - \frac{d}{R}\right) = \frac{g}{2}$$ $$1 - \frac{d}{R} = \frac{1}{2} \implies d = \frac{R}{2} = 3200 \text{ km}$$
Final Answer: $d = 3200$ km.
The force of gravitation between two bodies of mass $1$ kg each separated by $1$ m is:
The value of $g$ at the centre of Earth is:
The weight of a body at the equator is less than at the poles because:
At a height equal to Earth's radius, the value of $g$ is:
If Earth's radius decreased by $2\%$ keeping mass constant, $g$ would:
Gravitational Field and PotentialTopic 2
Gravitational Field Intensity ($\vec{E}_g$) at a point: force per unit mass placed at that point. $$\vec{E}_g = \frac{\vec{F}}{m} = -\frac{GM}{r^2}\hat{r}$$ SI unit: N/kg = m/s². It is a vector quantity directed toward the source mass.
Gravitational Potential ($V$): work done per unit mass in bringing a test mass from infinity to that point. $$V = -\frac{GM}{r}$$ SI unit: J/kg. Always negative (reference: $V = 0$ at infinity). Relationship: $\vec{E}_g = -\nabla V$, or in 1D: $E_g = -dV/dr$.
Gravitational Potential Energy (PE): $U = -\dfrac{Gm_1m_2}{r}$. Always negative. For two-particle system: $U = mV$.
Field/Potential of Common Bodies:
| Body | Field $E$ (outside) | Potential $V$ (outside) | At centre |
|---|---|---|---|
| Point mass / spherical shell | $GM/r^2$ | $-GM/r$ | Shell: $E=0, V=-GM/R$ |
| Solid sphere (inside, distance $r$) | $GMr/R^3$ | $-\dfrac{GM}{2R^3}(3R^2-r^2)$ | $V = -3GM/(2R)$ |
| Ring (axis, distance $x$) | $\dfrac{GMx}{(R^2+x^2)^{3/2}}$ | $-\dfrac{GM}{\sqrt{R^2+x^2}}$ | $E = 0, V = -GM/R$ |
Self-energy of uniform solid sphere: $U_{\text{self}} = -\dfrac{3GM^2}{5R}$.
Find the gravitational potential at the centre of a uniform thin spherical shell of mass $M$ and radius $R$.
Show solution
Inside a thin spherical shell, the field is zero everywhere. The potential is constant equal to its value at the surface: $$V_{\text{centre}} = V_{\text{surface}} = -\frac{GM}{R}$$
Final Answer: $V = -GM/R$.
Two particles of mass $m$ each are placed at $(0,0)$ and $(a,0)$. Find the work done in bringing a third particle of mass $m$ from infinity to the midpoint of the line joining them.
Show solution
Distance from each particle to midpoint = $a/2$. Potential at midpoint due to both: $$V = -\frac{Gm}{a/2} - \frac{Gm}{a/2} = -\frac{4Gm}{a}$$ Work done = $mV - 0 = -\dfrac{4Gm^2}{a}$ (negative, as gravity attracts).
Final Answer: $W = -4Gm^2/a$.
Gravitational potential at infinity is:
The gravitational field inside a uniform spherical shell is:
The gravitational potential at the centre of a uniform solid sphere of mass $M$, radius $R$:
The work done in bringing a body of mass $m$ from infinity to a point at distance $r$ from a mass $M$:
Two masses $M$ each are separated by $r$. Field intensity at the midpoint:
Planetary Motion and Satellites
Kepler's Laws and Orbital MotionTopic 1
Kepler's Laws:
- Law of Orbits: Each planet moves in an elliptical orbit with the Sun at one focus.
- Law of Areas: The line joining the planet to the Sun sweeps equal areas in equal times. This implies $L = m r^2 \dot\theta = \text{constant}$ (conservation of angular momentum).
- Law of Periods: $T^2 \propto a^3$, where $a$ is the semi-major axis. Specifically:
$$T^2 = \frac{4\pi^2}{GM}a^3$$
Orbital Velocity for circular orbit at height $h$: $$v_o = \sqrt{\frac{GM}{R+h}} = \sqrt{\frac{gR^2}{R+h}}$$ Near surface ($h \approx 0$): $v_o \approx \sqrt{gR} \approx 7.92$ km/s.
Orbital Period: $$T = 2\pi\sqrt{\frac{(R+h)^3}{GM}}$$
Areal velocity: $\dfrac{dA}{dt} = \dfrac{L}{2m} = \text{const}$.
For elliptical orbit with semi-major axis $a$, semi-minor axis $b$, eccentricity $e$:
- $r_{\min} = a(1-e)$ (perihelion), $r_{\max} = a(1+e)$ (aphelion)
- $v_{\max}$ at perihelion, $v_{\min}$ at aphelion; $v_p r_p = v_a r_a$
A satellite is orbiting Earth at a height equal to Earth's radius. Find its orbital velocity. ($g = 9.8$ m/s², $R = 6400$ km)
Show solution
$$v_o = \sqrt{\frac{gR^2}{R+h}} = \sqrt{\frac{gR^2}{2R}} = \sqrt{\frac{gR}{2}}$$ $$= \sqrt{\frac{9.8 \times 6.4 \times 10^6}{2}} = \sqrt{3.136 \times 10^7} \approx 5.6 \times 10^3 \text{ m/s} = 5.6 \text{ km/s}$$
Final Answer: $v_o \approx 5.6$ km/s.
The mean distance of Mars from the Sun is $1.524$ times that of Earth. Find its orbital period in years.
Show solution
By Kepler's third law: $T^2 \propto a^3$. $$\frac{T_M}{T_E} = \left(\frac{a_M}{a_E}\right)^{3/2} = (1.524)^{3/2}$$ $$= 1.524 \times \sqrt{1.524} = 1.524 \times 1.234 \approx 1.881$$ $T_M = 1.881 \times 1 \text{ year} \approx 1.88$ years.
Final Answer: $T_M \approx 1.88$ years.
According to Kepler's second law, the areal velocity of a planet:
If the distance of a planet from the Sun is doubled, its period of revolution becomes:
Orbital velocity of a satellite is independent of:
A planet's velocity at perihelion is $v_p$ and at aphelion is $v_a$. Then:
The ratio of orbital velocities of two satellites at heights $R$ and $3R$ above Earth's surface ($R$ = Earth's radius):
Escape Velocity, Satellite Energy and Geostationary SatellitesTopic 2
Escape Velocity: Minimum velocity needed to escape from a body's gravitational field to infinity. $$v_e = \sqrt{\frac{2GM}{R}} = \sqrt{2gR}$$ For Earth: $v_e \approx 11.2$ km/s. Relation: $v_e = \sqrt{2}\, v_o$ (escape velocity is $\sqrt{2}$ times orbital velocity at surface).
Satellite Energies (orbit radius $r = R+h$):
| Quantity | Formula |
|---|---|
| Kinetic energy | $K = \dfrac{1}{2}mv_o^2 = \dfrac{GMm}{2r}$ |
| Potential energy | $U = -\dfrac{GMm}{r}$ |
| Total energy | $E = K + U = -\dfrac{GMm}{2r}$ |
| Binding energy | $\|E\| = \dfrac{GMm}{2r}$ |
Total energy is negative, meaning the satellite is bound. To free it: supply binding energy.
Geostationary Satellite: Appears stationary above a point on Earth (used for communication). Properties:
- Orbits in equatorial plane
- Period $T = 24$ h (same as Earth's rotation)
- Height $h \approx 36000$ km above surface
- Orbital radius $\approx 42000$ km
- Orbital velocity $\approx 3.1$ km/s
Polar Satellite: Orbits in a plane passing over the poles; low altitude (~ 500-800 km); used for mapping, weather observation.
Weightlessness in satellite: Astronaut and satellite both have the same gravitational acceleration → apparent weight = 0.
Calculate the escape velocity from the surface of the Moon. Mass = $7.4 \times 10^{22}$ kg, radius = $1.74 \times 10^6$ m.
Show solution
$$v_e = \sqrt{\frac{2GM}{R}} = \sqrt{\frac{2 \times 6.67 \times 10^{-11} \times 7.4 \times 10^{22}}{1.74 \times 10^6}}$$ $$= \sqrt{\frac{9.87 \times 10^{12}}{1.74 \times 10^6}} = \sqrt{5.67 \times 10^6} \approx 2.38 \times 10^3 \text{ m/s} = 2.38 \text{ km/s}$$
Final Answer: $v_e \approx 2.38$ km/s.
A satellite of mass $200$ kg is in a circular orbit at height $400$ km above Earth's surface. Find the energy needed to take it to infinity. ($g = 9.8$, $R = 6400$ km)
Show solution
$$E_{\text{required}} = \text{Binding energy} = \frac{GMm}{2(R+h)} = \frac{gR^2 m}{2(R+h)}$$ $$= \frac{9.8 \times (6.4 \times 10^6)^2 \times 200}{2 \times 6.8 \times 10^6} = \frac{9.8 \times 4.096 \times 10^{13} \times 200}{1.36 \times 10^7}$$ $$= \frac{8.028 \times 10^{16}}{1.36 \times 10^7} \approx 5.9 \times 10^9 \text{ J}$$
Final Answer: $E \approx 5.9 \times 10^9$ J.
Escape velocity from Earth's surface is approximately:
Ratio of escape velocity to orbital velocity at Earth's surface:
The total energy of a satellite in orbit is:
Period of a geostationary satellite is:
If a satellite's orbital radius is doubled, its total energy:
Ready to test yourself?
Attempt the full timed mock tests — Main & Advanced level.
Start Mock Test 1 →